Post on 03-Jan-2016
Chapter 5Chapter 5
ZZ Transform Transform
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Z transform– Representation, analysis, and design of discrete signal
– Similar to Laplace transform
– Conversion of digital signal into frequency domain
1. Introduction1. Introduction
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z transform– Two-sided z transform
– One-sided z transform• If n < 0, x(n) = 0
2. 2. zz transform transform
0
( ) ( ) n
n
X z x n z
( )
( )
n
n
n
X z x
x n z
(5-1)
(5-2)
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Example 5-1(1) Non-causal
(2) Non-causal
5 4 3 2( ) ( ) 2 2 2n
n
X z x n z z z z z z
Fig. 5-1.
2 1 2( ) ( ) 2 2 2n
n
X z x n z z z z z
Fig. 5-1.
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Example 5-1(3) Causal
(4)
• Geometric series with common ratio of
Fig. 5-1.
Fig. 5-1.
1 2 3 4 5( ) ( ) 2 2 2n
n
X z x n z z z z z z
1, 0( )
0, 0
nx n
n
1 2
0
( ) ( ) 1n n
n n
X z x n z z z z
1z
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• Convergence of series
• Region of convergence
1 2
1
( ) 1
1 (1 ) ( 1)
X z z z
z z z
Fig. 5-2.
Region of convergence
(5-3)
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• Z=2
• Z=1/2
2 3( ) 1 1 2 (1 2) (1 2)
2 (2 1) 2
X z
2 3( ) 1 1 0.5 (1 0.5) (1 0.5)
1 2 4 8
X z
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Example 5-2
– z transform
0 , 0( )
, 0, 0n
nx n
b n b
1
0
( ) ( )
1 ( ) ,
n n
n
n
n
X z b z
b zz b
z z b
Region of convergence
Fig. 5-3.
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Characteristic of z transform
(1) is a polynomial equation of z and determined from samples, (2) can be reconstructed by removing in
(3) is independent to sampling interval,
(4) z transform of delayed signal by samples is
• z transform of delayed signal
• Expression of difference equation
( )X z ( )x n
( )x n nz ( )X z
( )X z T
m ( )mz X z
( ) ( )
( ) ( )m
x n X z
x n m z X z
{ ( )} ( ) n
n
z x n m x n m z
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•
(5) Same to discrete Fourier transform by replacing to
( )( ) ( )
( )
( )
n k m
n k
m k
k
m
x n m z x k z
z x k z
z X z
n m k
z j Te
( ) ( )j TsX j X e (5-
4)
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Table of z transform
Table 5-1.
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Example 5-3
(1)
(2)
(3)
( ) ( ), ( ) 1x n n X z
( ) ( )x n u n
0
1 2
11
( )
1
1, 1
1
n
n
X z z
z z
zz
( ) , 0anTx n e n
0
1 2 2
11
( )
1
1, 1
1
anT n
n
aT aT
aTaT
X z e z
e z e z
e ze z
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Ideally sampled function,
– Laplace transform
3. Relation between Z transform 3. Relation between Z transform and Laplace transformand Laplace transform
( )sx t
0
( ) ( ) ( )sn
x t x t t nT
0
00
00
0
( ) ( )
( ) ( )
( ) ( )
( )
sts s
st
n
st
n
nTs
n
X s x t e dt
x t t nT e dt
x t e t nT dt
x n e
(5-5)
(5-6)
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– z transform
– Relation
( ) ( ) sTe zX z X s
( ) ( ) n
n
X z x n z
(5-7)
(5-8)
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Example 5-4 (1)
(2)
( ) ( )x t u t1
( )X ss
( ) ( )x n u n0
11
( )
1, 1
1
n
n
X z z
zz
( ) ( )atx t e u t 1( )X s
s a
( ) ( )anTx n e u n0
11
( )
1, 1
1
anT n
n
aTaT
X z e z
e ze z
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(3) ( ) (sin ) ( )x t t u t2 2
( )X ss
( ) (sin ) ( )x n nT u n0
0
1
2 1
( ) (sin )
2
sin
2 cos 1
n
n
j nT j nTn
n
X z nT z
e ez
j
z T
z z T
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– Periodicity
– s-plane and z-plane
( )ss j m TsTe e
Fig. 5-4.
(5-9)
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Corresponding points(1) Left side plane in s plane inside of unit circle in z plane
(2) Right side plane in s plane out of unit circle
(3) axis in s plane unit circle in z plane
(4) Increased frequency in s plane mapped on the unit circle in z plane
j
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– Corresponding points
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Definition of inverse z transform
– Power series of
4. Inverse Z transform4. Inverse Z transform
1( ) ( )x n z X z
0
1 2 3
( ) ( )
(0) (1) (2) (3)
n
n
X z x n z
x x z x z x z
( )X z
1 20 1 2
1 20 1 2
( )N
NM
M
b b z b z b zX z
a a z a z a z
(5-10)
(5-11)
(5-12)
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– Three methods to obtain inverse z transform• Power series expansion
• Partial fraction expansion
• Residue
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– Power series expansion• Long division
1 20 1 2
1 20 1 2
1 2 3
( )
(0) (1) (2) (3)
NN
MM
b b z b z b zX z
a a z a z a z
x x z x z x z
(5-13)
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Example 5-5– Inverse z transform using long division
1 2
1 2
1 2( )
1 2
z zX z
z z
1 2 3
1 2 1 2
1 2
1 2
1 2 3
2 3
2 3 4
3 4
1 3 2 41 2 1 2
1 2
3
3 3 6
2 6
2 2 4
4 4
z z zz z z z
z z
z z
z z z
z z
z z z
z z
1 2 3( ) 1 3 2 4X z z z z
(0) 1, (1) 3, (2) 2, (3) 4,x x x x
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– Partial fraction expansion
1 20 1 2
1 20 1 2
( )N
NM
M
b b z b z b zX z
a a z a z a z
1 20 1 1 1
1 2
1 20
1 2
01
0 11
( )1 1 1
1
M
M
M
M
Mk
k k
Mk
k k
C C CX z B
p z p z p z
C z C z C zB
z p z p z p
C zB
z p
CB
p z
where is poles of ,
is coefficients for partial fraction, and
kp ( )X z
kC
0 / .N NB b a
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• Partial fraction for N>M
10 1
( )1
N M Mkr
rr k k
CX z B z
p z
where is calculated using long division.rB
1
( )( )
(1 ) ( )
k
k
k kz p
k z p
X zC z p
z
p z X z
1
mi
i k
D
z p
1 ( )( )
( )!k
m im
i km iz p
d X zD z p
m i dz z
(5-17)
(5-18)
(5-19)
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Example 5-6– Inverse z transform
1
1 2( )
1 0.25 0.375
zX z
z z
2( )
0.25 0.375 ( 0.75)( 0.5)
z zX z
z z z z
1 2( )( 0.75)( 0.5) 0.75 0.5
C z C zzX z
z z z z
1 2( )
( 0.75)( 0.5) 0.75 0.5
C CX z z
z z z z z z
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10.75 0.75
0.75
( 0.75) ( ) ( 0.75)
( 0.75)( 0.5)
10.8
0.5
z z
z
z X z zC
z z z
z
20.5 0.5
( 0.5) ( ) ( 0.5)
( 0.75)( 0.5)
0.8z z
z X z zC
z z z
0.8 0.8( )
0.75 0.5
z zX z
z z
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• Inverse z transform using table 5-1
1 0.80.8(0.75)
0.75nz
zz
1 0.80.8( 0.5)
0.5nz
zz
( ) 0.8 (0.75) ( 0.5) , 0n nx n n
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Example 5-7– Inverse z transform
• poles
1 2
1 2
1 2( )
1 0.5
z zX z
z z
2
2
2
1 2
( ) 2 1( )
( ) 0.5
2 1
( )( )
N z z zX z
D z z z
z z
z p z p
1 2
1
1 (1 4 0.5)0.5 0.5
2p j
*2 1 0.5 0.5p p j
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• Partial fraction
1
0 11 2 11
2
( )( ) ( ) ( )
z p
B z pz p X z C z pC
z z z p
1
21 1
11 2
( ) ( ) ( )( 2 1)
( )( )
2 1.5
0.5 0.5
0.5 3.5
z p
z p X z z p z zC
z z z p z p
j
j
j
*2 1 0.5 3.5C C j
0 1 2
1 1
( ) B C CX z
z z z p z p
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• z transform
• Inverse z transform
1 2
1 1
( ) 2
( 0.5 3.5) ( 0.5 3.5)2
0.5 0.5 0.5 0.5
C z C zX z
z p z p
j z j z
z j z j
1 1 2
1 1
2 3.5355(0.7071) cos(45 98.13 )
7.071(0.7071) cos(45 98.13 )
n
n
C z C zz n
z p z p
n
1(2) 2 ( )z n
( ) 2 ( ) 7.071(0.7071) cos(45 98.13 ), 0nx n n n n
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Example 5-8– Inverse z transform
2
2( )
( 0.5)( 1)
zX z
z z
1 22
( )
0.5 ( 1) ( 1)
D DX z C
z z z z
22
2
0.5
( 0.5)0.5 / (0.5 1) 2
( 0.5)( 1)z
z zC
z z z
2 2 2
1 2
1 1
21 1
( 1) ( ) ( 1)
( 0.5)( 1)
0.52
0.5 ( 0.5)
z z
z z
d z X z d z zD
dz z dz z z z
d z z z
dz z z
2 2 2
2 2
1 1
( 1) ( ) ( 1)2
( 0.5)( 1)z z
z X z z zD
z z z z
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• Inverse z transform
2
2 2 2( )
0.5 ( 1) ( 1)
z z zX z
z z z
( ) 2(0.5) 2 2 2 ( 1) (0.5) , 0n nx n n n n
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– Residue• Cauchy’s theory using contour integral
• Calculation of residue
11( ) ( )
2n
Cx n z X z dz
j
where contour integral including all poles.C
where ,
m is order of poles.
1( ) ( )nF z z X z
11( ) ( )
2n
Cx n z X z dz
j
1
1
1Residue ( ), ( ) ( )
( 1)! k
mm
k km z p
dF z p z p F z
m dz
(5-20)
(5-21)
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• For single pole
Unit circle 1z
Fig. 5-5.
1
Residue ( ), ( ) ( )
( ) ( )k
k k
nk z p
F z p z p F z
z p z X z
(5-22)
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Example 5-9– Find discrete time signal
• If ,
• Inverse z transform
( )( 0.75)( 0.5)
zX z
z z
1
( )( 0.75)( 0.5)
( 0.75)( 0.5)
n
n
z zF z
z z
z
z z
( ) Residue ( ),0.75 Residue ( ), 0.5x n F z F z
1( ) ( )nF z z X z
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• Sum of residue
0.75
0.75
Residue ( ),0.75 ( 0.75) ( )
( 0.75)
( 0.75)( 0.5)
(0.75)
0.75 0.5
0.8(0.75)
z
n
z
n
n
F z z F z
z z
z z
0.5
0.5
Residue ( ), 0.5 ( 0.5) ( )
( 0.5)
( 0.75)( 0.5)
0.8( 0.5)
z
n
z
n
F z z F z
z z
z z
( ) 0.8 (0.75) ( 0.5)n nx n
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Example 5-10– Inverse z transform
where and as 1 0.5 0.5p j 1 0.5 0.5p j *2 1 .p p
2
2
2 1( )
0.5
z zX z
z z
2
1 2
2 1( )
( )( )
z zX z
z p z p
1 2 21
2 2
( 2 1) ( 2 1)( ) ( )
0.5 ( 0.5)
n nn z z z z z z
F z z X zz z z z z
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Fig. 5-5.
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• n=0,2
2
2 1( )
( 0.5)
z zF z
z z z
1 2(0) Residue ( ),0 Residue ( ), Residue ( ),x F z F z p F z p
0
2
2
0
Residue ( ),0 ( )
( 2 1)
( 0.5)
1/ 0.5 2
z
z
F z zF z
z z z
z z z
1
1
1 1
21
1 2
Residue ( ), ( ) ( )
( )( 2 1)
( )( )
0.5 3.5
z p
z p
F z p z p F z
z p z z
z z p z p
j
2Residue ( ), 0.5 3.5F z p j
1 2(0) Residue ( ),0 Residue ( ), Residue ( ),
2 0.5 3.5 0.5 3.5
1
x F z F z p F z p
j j
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• n>0
2
2
( 2 1)( )
( 0.5)
nz z zF z
z z z
1 2( ) Residue ( ), Residue ( ),x n F x p F z p
1
1
1 1
21
1 2
Residue ( ), ( ) ( )
( ) ( 2 1)
( )( )
z p
n
z p
F z p z p F z
z p z z z
z z p z p
45 98.131Residue ( ), (0.7071 ) (3.5355 )
3.5355(0.7071) cos(45 98.13 ) sin(45 98.13 )
j n j
n
F z p e e
n j n
2Residue ( ), 3.5355(0.7071) cos(45 98.13 ) sin(45 98.13 )nF z p n j n
1 2( ) Residue ( ), Residue ( ),
7.071(0.7071) cos(45 98.13 ), 0n
x n F z p F z p
n n
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Example 5-11
where F(z) has poles at z=0.5, and z=1.
2
2( )
( 0.5)( 1)
zX z
z z
( ) Residue ( ), kx n F z p
11
2( ) ( )
( 0.5)( 1)
nn z
F z z X zz z
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• Sum of residue
( ) 2 ( 1) (0.5)nx n n
2 1
2
1
1
2
1
2
( 1)Residue ( ),1
( 0.5)( 1)
( 0.5)( 1)
( 0.5)
0.5( 1) 1 / (0.5) 2( 1)
n
z
n n
z
d z zF z
dz z z
z n z z
z
n n
1 2( ) Residue ( ), Residue ( ),x n F z p F z p
1 1
2 2
0.5
2
( 0.5)Residue ( ),0.5)
( 0.5)( 1) ( 1)
0.5(0.5) / (0.5) 2(0.5)
n n
z
n n
z z zF z
z z z
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linearity
Convolution
Differentiation
5. Characteristic of z transform5. Characteristic of z transform
1 2 1 2( ) ( ) ( ) ( )ax n bx n aX z bX z
( ) ( ) ( )k
y n h k x n k
( ) ( ) ( )Y z H z X z
( ) ( )
( )( )
x n X z
dX znx n z
dz
(5-23)
(5-24)
(5-25)